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Sampling With The Normal Distribution

  1. Feb 9, 2010 #1
    I have been puzzling over this problem for about a week now and cannot find the answer. In my opinion it is very theoretical, but I know I ma not the best mathematician on here so maybe someone else could look at this.

    Thank you.
  2. jcsd
  3. Feb 9, 2010 #2
    You need to derive the probability distribution for [tex]\bar{X}[/tex]. This is possible, since it is a sum of n quantities for which you know the distribution. Tthe distribution for [tex]\bar{X}[/tex] willl depend on n. You can then calculate the probability that [tex]|\bar{X}-\mu|>\sigma/2[/tex] in the ordinary way. This gives you an inequality for n. Then find the smallest n that satisfies the inequality.

    When you know n, you have completely specified the distribution for [tex]\bar{X}[/tex], and you can do part (b).

    So the first step is to find the distribution of a sum of quantities, expressed in terms of their individual distributions.

  4. Feb 9, 2010 #3
    Do you mean [tex]\bar{X}[/tex]~N([tex]\mu[/tex],[tex]\sigma[/tex]2/n)

    Where do I go from here if I knew the probability then I think I could manage.
  5. Feb 9, 2010 #4
    Cool, that's the same thing I got. From that you can write the explicit gaussian probability density for e.g. [tex]u:=\bar{X}-\mu[/tex]. Call it e.g. [tex]\rho_n(u)[/tex]. It will depend on n of course. It will simply the usual normalied gaussian centered around u=0, with variance [tex]\sigma^2/n[/tex] I think.

    Then the probability [tex]P(n)[/tex] that [tex]|u|>\sigma/2[/tex] is given as a integral that you can solve using the error function:

    P(n) = \int_{-\infty}^{-\sigma/2} \rho_n(u)du + \int^{\infty}_{\sigma/2} \rho_n(u)du

    Then find the smallest n such that P(n)>0.05. Then fix this value of n, and just do another simple integral to get the answer to part b.


  6. Feb 9, 2010 #5
    Got it now. It comes out n>16 so at least n=16. I used a different method which I will post now.
  7. Feb 9, 2010 #6


    =[tex]\phi[/tex]([tex]\sqrt{n}[/tex]/2) - (1-([tex]\phi[/tex]([tex]\sqrt{n}[/tex]/2))>0.95








    But n is integer
    Therefore n>16

    The minimum value of n is therefore 16
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